A mixture of carbon dioxide and an unknown gas was allowed to effuse from a container.?

Question

The carbon dioxide took 1.25 times as long to escape as the unknown gas. Which one could be the unknown gas?
A) Cl2 =71.0 g/mol
B) CO = 28.01 g/mol
C) HCl = 36.46 g/mol
D) H2
E) SO2 = 64.06 g/mol

Can someone explain how to figure this equation. I found all the g/mol except for H2. I do not understand how to figure this equation.

Answer 1

I'm going with the carbon monoxide.

Use Graham's Law of Effusion: r1/r2 = sqrt(M2/M1) where r1 and r2 are the rates of effusion and M2 and M1 are the molar masses.

Let's call CO2 #2. Since the CO2 took LONGER, it's easier to put that one in the demoninator. So, now we have

rx/rCO2 = sqrt(44/M1)

Since we know the CO2 takes 1.25 times longer, that tells us that rx/rCO2 = 1.25. Plugging that in, we get

1.25=sqrt(44/M1) Now, square both sides.
1.5625 = 44/M1 Multiply both sides by M1.
M1 x 1.5625 = 44 Divide both sides by 1.5625.
M1 = 28.16 g/mol.

Now, look back at your molar masses. I realize that 28.16 isn't the exact molar mass of CO, but in my experience, it's not unusual to have small rounding errors with problems like this.

So, like I said, I'll vote for carbon monoxide. If anyone sees any errors in my work, please let me know. I have an exam on this stuff on Monday!

Answer 2

The rate at which a gas diffuses is inversely proportional to the square root of its molecular weight. CO2 has a MW of 44, so the unknown gas has a MW of (SQR 0.8)times 44 = 39.4, so I'd go for HCl as the closest. H2 is only 2 g/mol.

Answer 3

the effusion rate will be a function of the gas density

since the carbon dioxide took longer to effuse, it is heavier than the other gas, but not much heavier, in fact, it is roughly 1.25 times heavier

look up the density of CO2 and see which gas fits the bill, don't worry about hydrogen, you know it is much lighter than those other gasses (but you can look them all up on a periodic table of the elements)

Answer 4

you pick to remodel all grams to moles. i am going to apply CO2: 12 g / 40 4 g/mol = 0.273 mol Do an similar for N2 and H2O. once you've finished that, upload up the three mole values to get the entire moles of gasoline contained in the sphere. i am going to apply a million.0 mol because the entire for my very last element. a million.0 likely isn't the most ideal answer, it is only a accessible determination. What you do is divide the moles of each and every gasoline by technique of the entire moles, then multiply that value by technique of one hundred ten. So, utilising my celebration, i'd do 0.273 / a million.0 = 0.273 Then one hundred ten x 0.273 to get the perfect answer So, get the entire quantity and placed it contained in the denominator the position I used a million.0 and also you're off to the races!

Answer 5

H2 is 2 g/mol

Answer 6

H = 1.01 g
H2 = (1.01 x 2) g = 2.02 g
this is because there are 2 Hydrogen atoms in the Hydrogen Gas!

Answer 7

CO